In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is $\frac{5}{6}$. What is the probability that he will knock down fewer than 2 hurdles?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{5^{10}}{2×6^9}$
Given, probability of clearing a hurdle = $\frac{5}{6}$
Let E be the event 'knock down a hurdle' then
$\text{p = P(E) = 1 - P(clearing a hurdle)} = 1-\frac{5}{6}=\frac{1}{6}$, so $q = 1 − p =1-\frac{1}{6}=\frac{5}{6}$
As a player has to cross 10 hurdles, so there are 10 trials i.e. $n = 10$.
Thus, we have a binomial distribution with $p =\frac{1}{6},q=\frac{5}{6}$ and $n = 10$.
Required probability = P(knocking down less than 2 hurdles) = P(0) + P(1)
$={^{10}C}_0q^{10}+{^{10}C}_1pq^9=(q+10p)q^9$
$=\left(\frac{5}{6}+10.\frac{1}{6}\right)\left(\frac{5}{6}\right)^9=\frac{5}{2}\left(\frac{5}{6}\right)^9=\frac{5^{10}}{2×6^9}$